PYQ NEET- Magnetism And Matter L-3

Question: A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now the torque required to keep the magnet in this new position is#

A) $\frac{W}{\sqrt{3}}$

B) $\sqrt{3} \mathrm{~W}$

C) $\frac{\sqrt{3} W}{2}$

D) $\frac{2 W}{\sqrt{3}}$

Answer: $\sqrt{3} \mathrm{~W}$#

Sol:#

$\because$ Work done in rotating the magnet $$ W=M B\left(\cos \theta_0-\cos \theta\right) $$

Where, $M=$ magnetic moment of the magnet $$ \begin{aligned} B & =\text { magnetic field } \ W & =M B\left(\cos 0^{\circ}-\cos 60^{\circ}\right) \ & =M B\left(1-\frac{1}{2}\right)=\frac{M B}{2} \end{aligned} $$ $\therefore \quad M B=2 W$ Torque on a magnet in this position is given by,

$\begin{aligned} \tau & =\mathbf{M} \times \mathbf{B} \ & =M B \cdot \sin \theta=2 W \cdot \sin 60^{\circ} \ & =2 W \frac{\sqrt{3}}{2}=W \sqrt{3} \end{aligned}$